Choice D is correct. The given margin of error of 3% indicates that the actual percent of all US teens who are heavy texters is likely within 3% of the estimate of 30%, or between 27% and 33%. Therefore, it is unlikely, or doubtful, that the percent of all US teens who are heavy texters would be 35%.

Choice A is incorrect. The margin of error doesn’t provide any information about the accuracy of reporting in the study. Choice B is incorrect. Based on the estimate and given margin of error, it is unlikely that the percent of all US teens who are heavy texters would be less than 27%, but it is possible. Choice C is incorrect. While the percent of all US teens who are heavy texters is likely between 27% and 33%, any value within this interval is equally likely. We can’t be certain that the value is exactly 33%.

The correct answer is $\mathrm{2,520}$. There are $60$ minutes in one hour. At a rate of $42$ posters per minute, the number of posters produced in one hour can be determined by $\left(\frac{42 \text{posters}}{1 \text{minute}}\right)\left(\frac{60 \text{minutes}}{1 \text{hour}}\right)$, which is $\mathrm{2,520}$ posters per hour.

The correct answer is . It is given that no contestant received the same score on two different days, so each of the contestants who received a score of 5 is represented in the “5 out of 5” column of the table exactly once. Therefore, the probability of selecting a contestant who received a score of 5 on Day 2 or Day 3, given that the contestant received a score of 5 on one of the three days, is found by dividing the total number of contestants who received a score of 5 on Day 2 or Day 3 by the total number of contestants who received a score of 5, which is given in the table as 7. So the probability is . Note that 5/7, .7142, .7143, and 0.714 are examples of ways to enter a correct answer.

Choice A is correct. The total number of possible outcomes for rolling a fair $14$-sided die is $14$. The number of possible outcomes for rolling a $2$ is $1$. The probability of rolling a $2$ is the number of possible outcomes for rolling a $2$ divided by the total number of possible outcomes, or $\frac{1}{14}$.

Choice B is incorrect. This is the probability of rolling a number no greater than $2$.

Choice C is incorrect. This is the probability of rolling a number greater than $2$.

Choice D is incorrect. This is the probability of rolling a number other than $2$.

Choice A is correct. It’s given that $20\%$ of the students surveyed responded that they intend to enroll in the study program. Therefore, the proportion of students in Spanish club who intend to enroll in the study program, based on the survey, is $0.20$. Since there are $55$ total students in Spanish club, the best estimate for the total number of these students who intend to enroll in the study program is $55\left(0.20\right)$, or $11$.

Choice B is incorrect. This is the best estimate for the percentage, rather than the total number, of students in Spanish club who intend to enroll in the study program.

Choice C is incorrect. This is the best estimate for the total number of Spanish club students who do not intend to enroll in the study program.

Choice D is incorrect. This is the total number of students in Spanish club.

Choice B is correct. It’s given that each serving of Crunchy Grain cereal provides 5% of an adult’s daily allowance of potassium, so x servings would provide x times 5%. The percentage of an adult’s daily allowance of potassium, p, is 5 times the number of servings, x. Therefore, the percentage of an adult’s daily allowance of potassium can be expressed as .

Choices A, C, and D are incorrect and may result from incorrectly converting 5% to its decimal equivalent, which isn’t necessary since p is expressed as a percentage. Additionally, choices C and D are incorrect because the context should be represented by a linear relationship, not by an exponential relationship.

Choice D is correct. Let $n\%$ represent the percent by which the positive quantity $x$ is decreased to result in $0.35x$. The value of $n$ can be found by solving the equation $x-\left(\frac{n}{100}\right)x=0.35x$. Since $x$ is a common factor of each of the terms on the left-hand side of this equation, the equation can be rewritten as $x\left(1-\frac{n}{100}\right)=0.35x$. Dividing each side of this equation by $x$ yields $1-\frac{n}{100}=0.35$. Multiplying each side of this equation by $100$ yields $100-n=35$. Subtracting $100$ from each side of this equation yields $-n=-65$. Dividing each side of this equation by $-1$ yields $n=65$. Therefore, the expression $0.35x$ represents the result of decreasing the positive quantity $x$ by $65\%$.

Choice A is incorrect. Decreasing the quantity $x$ by $3.5\%$ yields $x-0.035x$, or $0.965x$, not $0.35x$.

Choice B is incorrect. Decreasing the quantity $x$ by $35\%$ yields $x-0.35x$, or $0.65x$, not $0.35x$.

Choice C is incorrect. Decreasing the quantity $x$ by $6.5\%$ yields $x-0.065x$, or $0.935x$, not $0.35x$.

Choice B is correct. If a data set contains an even number of data values, when the data values are listed in ascending or descending order, the median is between the two middle values. The given data set contains $8$ values. When listed in ascending order, the data set is $4$, $4$, $4$, $5$, $5$, $6$, $10$, $18$ and the two middle values are $5$ and $5$. Since the two middle values are the same, the median must be $5$.

Choice A is incorrect. This value is between the two middle values in the list shown, not the two middle values when the data values are listed in ascending or descending order.

Choice C is incorrect. This is the mean, not the median, of the data set.

Choice D is incorrect. This is the range, not the median, of the data set.

Choice A is correct. It’s given that there are 2 hybrid cars priced at no more than $25,000. It’s also given that there are 14 cars total for sale. Therefore, the probability of selecting a hybrid priced at no more than $25,000 when one car is chosen at random is .

Choice B is incorrect. This is the probability of selecting a hybrid car priced greater than $25,000 when choosing one car at random. Choice C is incorrect. This is the probability, when choosing randomly from only the hybrid cars, of selecting one priced at no more than $25,000. Choice D is incorrect. This is the probability of selecting a hybrid car when selecting at random from only the cars priced greater than $25,000.

Choice D is correct. Since 1 minute = 60 seconds and 1 hour = 60 minutes, it follows that 1 hour = (60)(60), or 3,600 seconds. Using this conversion factor, the space station’s average speed of 4.76 miles per second is equal to an average speed of , or 17,136 miles per hour.

Choice A is incorrect. This is the space station’s average speed in miles per minute. Choice B is incorrect. This is double the space station’s average speed in miles per minute, or the number of miles the space station travels on average in 2 minutes. Choice C is incorrect. This is triple the space station’s average speed in miles per minute, or the number of miles the space station travels on average in 3 minutes. 

Choice A is correct. The data point with has a y-coordinate of 12. The y-value predicted by the line of best fit at is approximately 11. The difference between the y-coordinate of the data point and the y-value predicted by the line of best fit at is , or 1.

Choices B and C are incorrect and may result from incorrectly reading the scatterplot. Choice D is incorrect. This is the y-coordinate of the data point at .

Choice D is correct. It’s given that the ratio of m to k is estimated to be 1 to 5. Therefore, when , the relationship between these ratios can be expressed by the proportion . Multiplying both sides of this equation by 25 yields .

Choices A, B, and C are incorrect and may result from calculation errors.

Choice D is correct. The increase in Iceland’s population can be found by multiplying the increase in population density, in people per square kilometer, by the area, in square kilometers. It’s given that the population density of Iceland was 2.5 people per square kilometer in 1990 and 3.3 people per square kilometer in 2014. The increase in population density can be found by subtracting 2.5 from 3.3, which yields 0.8. It’s given that the land area of Iceland was 100,250 square kilometers. Thus, the increase in population is , or 80,200.

Alternate approach: It’s given that the population density of Iceland, in people per square kilometer of land area, in 1990 was 2.5. Since the land area of Iceland was 100,250 square kilometers, it follows that the population of Iceland in 1990 was , or 250,625. Similarly, the population of Iceland in 2014 was , or 330,825. The population increase is the difference in the population from 1990 to 2014, or , which yields 80,200. Therefore, Iceland’s population increased by 80,200 from 1990 to 2014.

Choice A is incorrect. This is the population of Iceland in 2014. Choice B is incorrect and may result from dividing 3.3 by 2.5, instead of subtracting 2.5 from 3.3. Choice C is incorrect and may result from dividing the population of Iceland in 1990 by 2.

The correct answer is 94. Of those interviewed, “strongly favor” or “somewhat favor” the use of nuclear energy, and interviewees “somewhat oppose” or “strongly oppose” the use of nuclear energy. The difference between the sizes of the two surveyed groups is . The proportion of this difference among the entire group of interviewees is . Because the sample of interviewees was selected at random from US residents, it is reasonable to assume that the proportion of this difference is the same among all US residents as in the sample. Therefore, the best estimate, in millions, of the difference between the number of US residents who somewhat favor or strongly favor the use of nuclear energy and the number of those who somewhat oppose or strongly oppose it is , which to the nearest million is 94.

Choice B is correct. When a data set has an odd number of values, the median can be found by ordering the values from least to greatest and determining the value in the middle. Since the values are already presented in order from least to greatest and there are 7 values, the median is the fourth value in the list. Therefore, the median is 3.

Choice A is incorrect. This is the mode. Choice C is incorrect. This is the mean. Choice D is incorrect. This is the range.

Choice C is correct. The median of a data set represented in a box plot is represented by the vertical line within the box. It follows that the median mass of the gazelles in group $1$ is $25$ kilograms, and the median mass of the gazelles in group $2$ is $24$ kilograms. Since $25$ kilograms is greater than $24$ kilograms, the median mass of group $1$ is greater than the median mass of group $2$.

Choice A is incorrect. The mean mass of each of the two groups cannot be determined from the box plots.

Choice B is incorrect. The mean mass of each of the two groups cannot be determined from the box plots.

Choice D is incorrect and may result from conceptual or calculation errors.

The correct answer is 8. The 6 students represent of those invited to join the committee. If x people were invited to join the committee, then . Thus, there were people invited to join the committee. It follows that there were teachers and school and district administrators invited to join the committee. Therefore, there were 8 more teachers than school and district administrators invited to join the committee.

The correct answer is $980$. It's given that the ratio $140$ to $m$ is equivalent to the ratio $4$ to $28$. Therefore, the value of $m$ can be found by solving the equation $\frac{140}{m}=\frac{4}{28}$. Multiplying each side of this equation by $m$ yields $140=\frac{4m}{28}$. Multiplying each side of this equation by $28$ yields $\mathrm{3,920}=4m$. Dividing each side of this equation by $4$ yields $980=m$. Therefore, the value of $m$ is $980$.

The correct answer is $9$. It’s given that the customer spent $\$27$ to purchase oranges at $\$3$ per pound. Therefore, the number of pounds of oranges the customer purchased is $\$27\left(\frac{1 \text{pound}}{\$3}\right)$, or $9$ pounds.

The correct answer is $180$. It’s given that a mechanical device produces items at a constant rate of $60$ items per hour. This rate can be written as $\frac{60 \text{items}}{1 \text{hour}}$. Let $x$ represent the number of items the mechanical device will produce in $3$ hours at the given rate. It follows that $\frac{60 \text{items}}{1 \text{hour}}=\frac{x \text{items}}{3 \text{hours}}$, which can be written as $\frac{60}{1}=\frac{x}{3}$, or $60=\frac{x}{3}$. Multiplying each side of this equation by $3$ yields $180=x$. Therefore, at the given rate, the mechanical device will produce $180$ items in $3$ hours.

Alternate approach: It's given that a mechanical device produces items at a constant rate of $60$ items per hour. At this rate, the mechanical device will produce $\left(\frac{60 \text{items}}{1 \text{hour}}\right)\left(3 \text{hours}\right)$, or $180$ items in $3$ hours.

Choice B is correct. Selecting a sample of a reasonable size at random to use for a survey allows the results from that survey to be applied to the population from which the sample was selected, but not beyond this population. In this case, the population from which the sample was selected is all fourth-grade students at a certain school. Therefore, the results of the survey can be applied to all fourth-grade students at the school.

Choice A is incorrect. The results of the survey can be applied to the 40 students who were surveyed. However, this isn’t the largest group to which the results of the survey can be applied. Choices C and D are incorrect. Since the sample was selected at random from among the fourth-grade students at a certain school, the results of the survey can’t be applied to other students at the school or to other fourth-grade students who weren’t represented in the survey results. Students in other grades in the school or other fourth-grade students in the country may feel differently about announcements than the fourth-grade students at the school.

Choice B is correct. The graphical model that most closely fits the data in the scatterplot is a model in which the number of data points above and below the model are approximately balanced. Fitting a graphical model to the data shown results in an upward-facing parabola with a y-intercept near and a vertex with an approximate x-value of 2.5. Of the given choices, only choice B gives an equation of an upward-facing parabola with a y-intercept at . Furthermore, substituting 2.5 for x into the equation in choice B yields . This is approximately the y-value of the vertex of the model.

Choices A, C, and D are incorrect. These equations don’t give a graphical model that best fits the data. At , they have y-values of , , and 3, respectively. At , they have y-values of , , and 3, respectively.

Choice A is correct. The purpose of surveying a random sample of residents is to approximate the percent of the town residents that are satisfied with the concession stand. The sample doesn’t necessarily get the same result as surveying every resident of the town, nor would another sample necessarily have identical results. Therefore, although it’s possible that either statement I or statement II could prove true by surveying every resident of the town, these statements cannot be proven true solely based on the results of the sample.

Choice B is incorrect because surveying a sample of the town residents may not have the same result as surveying all the town residents. Choices C and D are incorrect because surveying a different sample of residents could yield different results.

Choice A is correct. The average rate of change in temperature of the coffee in degrees Fahrenheit per minute is calculated by dividing the difference between two recorded temperatures by the number of minutes in the corresponding interval of time. Since the time intervals given are all 10 minutes, the average rate of change is greatest for the points with the greatest difference in temperature. Of the choices, the greatest difference in temperature occurs between 0 and 10 minutes.

Choices B, C, and D are incorrect and may result from misinterpreting the average rate of change from the graph.

The correct answer is $\mathrm{2,432}$. It's given that $4 \text{cups}=1 \text{quart}$. It follows that $76$ quarts is equivalent to $\left(76 \text{quarts}\right)\left(\frac{4 \text{cups}}{1 \text{quart}}\right)$, or $304$ cups. It's also given that $8 \text{fluid ounces}=1 \text{cup}$. It follows that $304$ cups is equivalent to $\left(304 \text{cups}\right)\left(\frac{8 \text{fluid ounces}}{1 \text{cup}}\right)$, or $\mathrm{2,432}$ fluid ounces.

Choice A is correct. The mode is the data value with the highest frequency. So for the data shown, the mode is 18. The median is the middle data value when the data values are sorted from least to greatest. Since there are 20 ages ordered, the median is the average of the two middle values, the 10th and 11th, which for these data are both 19. Therefore, the median is 19. The mean is the sum of the data values divided by the number of the data values. So for these data, the mean is .

Since the mode is 18, the median is 19, and the mean is 20, .

Choices B and D are incorrect because the mean is greater than the median. Choice C is incorrect because the median is greater than the mode.

Alternate approach: After determining the mode, 18, and the median, 19, it remains to determine whether the mean is less than 19 or more than 19. Because the mean is a balancing point, there is as much deviation below the mean as above the mean. It is possible to compare the data to 19 to determine the balance of deviation above and below the mean. There is a total deviation of only 6 below 19 (the 6 values of 18); however, the data value 30 alone deviates by 11 above 19. Thus the mean must be greater than 19.

Choice D is correct. If a marble is chosen at random from the bag, the probability of choosing a marble of a certain color is the number of marbles of that color divided by the total number of marbles in the bag. Since there are 10 blue marbles in the bag, and there are 40 total marbles in the bag, the probability that the marble chosen will be blue is .

Choices A, B, and C are incorrect. These represent the probability that the marble chosen won’t be blue (choice A), will be green (choice B), and won’t be green (choice C).

Choice D is correct. It's given that the bar graph summarizes the charge, in kilowatt-hours $\left(\text{kWh}\right)$, a battery received each day for $15$ days. The height of each bar in the bar graph shown represents the number of days the battery received the charge, in $\text{kWh}$, specified at the bottom of the bar. The bar for a charge of $0 \text{kWh}$ reaches a height of $6$. Therefore, the battery received a charge of $0 \text{kWh}$ for $6$ of these days.

Choice A is incorrect. This is the charge, in $\text{kWh}$, that the battery received, not the number of days the battery received this charge.

Choice B is incorrect. This is the number of days the battery received a charge of either $8$, $16$, or $23$ $\text{kWh}$.

Choice C is incorrect. This is the number of days the battery received a charge of $11$ $\text{kWh}$.

Choice D is correct. The total number of gas station customers on Tuesday was 135. The table shows that the number of customers who did not purchase gasoline was 50. Finding the ratio of the number of customers who did not purchase gasoline to the total number of customers gives the probability that a customer selected at random on that day did not purchase gasoline, which is .

Choice A is incorrect and may result from finding the probability that a customer did not purchase a beverage, given that the customer did not purchase gasoline. Choice B is incorrect and may result from finding the probability that a customer did not purchase gasoline, given that the customer did not purchase a beverage. Choice C is incorrect and may result from finding the probability that a customer did purchase a beverage, given that the customer did not purchase gasoline.

Choice A is correct. A frequency table is a table that lists the data value and shows the number of times the data value occurs. In the data listed, the number $4$ occurs four times, the number $8$ occurs three times, and the number $13$ occurs two times. This corresponds to the table in choice A.

Choice B is incorrect. This table has the values for number and frequency reversed.

Choice C is incorrect because the frequency values don't represent the data listed.

Choice D is incorrect. This table represents the listed number values as the frequency values.

Choice C is correct. If the butterfly traveled 2,100 miles in 120 days, then it traveled, on average, miles per day.

Choice A is incorrect. This is approximately the average amount of time, in days, it took the butterfly to fly one mile: days per mile. Choice B is incorrect and may result from an arithmetic error. Choice D is incorrect. This is the number of hours in a day rather than the number of miles flown per day.

The correct answer is $14$. The maximum value is the largest value in the data set. The frequency refers to the number of times a data value occurs. The given frequency table shows that for this data set, the data value $6$ occurs three times, the data value $7$ occurs three times, the data value $8$ occurs eight times, the data value $9$ occurs eight times, the data value $10$ occurs nine times, the data value $11$ occurs eleven times, the data value $12$ occurs nine times, the data value $13$ occurs zero times, and the data value $14$ occurs six times. Therefore, the maximum data value in the data set is $14$.

Choice D is correct. The table shows that there are a total of 16 kittens that have a chocolate-colored coat. Of the 16 with a chocolate-colored coat, 12 have deep blue eyes. Therefore, the fraction of chocolate-colored kittens with deep blue eyes is simply the ratio of those two numbers, or .

Choice A is incorrect; this is the fraction of all chocolate-colored kittens. Choice B is incorrect; this is the fraction of kittens with deep blue eyes that have a chocolate-colored coat. Choice C is incorrect; this is the fraction of cream-tortoiseshell-colored kittens with deep blue eyes.

The correct answer is $24$. The equation $\frac{24x}{ny}=4$ can be rewritten as $\left(\frac{24}{n}\right)\left(\frac{x}{y}\right)=4$. It's given that $\frac{x}{y}=4$. Substituting $4$ for $\frac{x}{y}$ in the equation $\left(\frac{24}{n}\right)\left(\frac{x}{y}\right)=4$ yields $\left(\frac{24}{n}\right)\left(4\right)=4$. Multiplying both sides of this equation by $n$ yields $\left(24\right)\left(4\right)=4n$. Dividing both sides of this equation by $4$ yields $24=n$. Therefore, the value of $n$ is $24$.

Choice B is correct. The line of best fit shown intersects the y-axis at a positive y-value and has a negative slope. The graph of an equation of the form $y=a+bx$, where $a$ and $b$ are constants, intersects the y-axis at a y-value of $a$ and has a slope of $b$. Of the given choices, only choice B represents a line that intersects the y-axis at a positive y-value, $13.5$, and has a negative slope, $-0.8$.

Choice A is incorrect. This equation represents a line that has a positive slope, not a negative slope.

Choice C is incorrect. This equation represents a line that intersects the y-axis at a negative y-value, not a positive y-value, and has a positive slope, not a negative slope.

Choice D is incorrect. This equation represents a line that intersects the y-axis at a negative y-value, not a positive y-value.

The correct answer is $110$. Let $x$ represent the total number of items in the sample. It’s given that $80\%$ of the items are faulty and that there are $88$ faulty items in the sample. Therefore, $80\%$ of $x$ is $88$. Since $80\%$ can be rewritten as $\frac{80}{100}$, it follows that $\frac{80}{100}x=88$. Multiplying both sides of this equation by $100$ yields $80x=\mathrm{8,800}$. Dividing both sides of this equation by $80$ yields $x=110$. Therefore, there are $110$ total items in the sample.

Choice B is correct. The line of best fit passes through the point . Therefore, the line of best fit predicts a temperature of 39°F for a location in Lake Tahoe Basin with an elevation of 8,500 feet.

Choice A is incorrect. This is the lowest temperature listed on the scatterplot, and the line of best fit never crosses this value for any of the elevations shown. Choice C is incorrect. According to the line of best fit, the temperature of 41°F is predicted for an elevation of slightly greater than 7,500 feet, not an elevation of 8,500 feet. Choice D is incorrect. According to the line of best fit, the temperature of 43°F is predicted for an elevation of roughly 6,700 feet, not an elevation of 8,500 feet.

Choice B is correct. According to the table, 140 students from the two high schools completed summer internships in 2010. Of these, 65 were from Valley High School. Therefore, of the students who completed summer internships in 2010, represents the fraction who were from Valley High School.

Choice A is incorrect. This is the difference between the numbers of students from the two high schools who completed internships in 2010 divided by the total number of students from the two schools who completed internships that year. Choice C is incorrect. This is the fraction of students from Foothill High School who completed internships out of all the students who completed internships in 2010. Choice D is incorrect. This is the number of students from Valley High School who completed internships in 2010 divided by the number of students from Foothill High School who completed internships in 2010.

Choice D is correct. The proportion of the paper clips that are size large can be written as $\frac{\mathrm{234,000}}{\mathrm{300,000}}$, or $0.78$. Therefore, the percentage of the paper clips that are size large is $0.78\left(100\right)$, or $78\%$.

Choice A is incorrect. This is the percentage of the paper clips that are not size large.

Choice B is incorrect and may result from conceptual or calculation errors.

Choice C is incorrect and may result from conceptual or calculation errors.

Choice D is correct. It’s given that $a$ is $\mathrm{2,241}\%$ of the sum of $b$ and $c$. This can be represented by the equation $a=\left(\frac{\mathrm{2,241}}{100}\right)\left(b+c\right)$, or $a=22.41\left(b+c\right)$. It’s also given that that $b$ is $83\%$ of $c$. This can be represented by the equation $b=\left(\frac{83}{100}\right)c$, or $b=0.83c$. Dividing both sides of this equation by $0.83$ yields $\frac{b}{0.83}=c$. Substituting $\frac{b}{0.83}$ for $c$ in the equation $a=22.41\left(b+c\right)$ yields $a=22.41\left(b+\frac{b}{0.83}\right)$, or $a=22.41\left(\frac{1.83b}{0.83}\right)$, which is equivalent to $a=\frac{41.0103b}{0.83}$, or $a=49.41b$. This equation is equivalent to $a=\left(\frac{\mathrm{4,941}}{100}\right)b$; therefore, $a$ is $\mathrm{4,941}\%$ of $b$.

Choice A is incorrect and may result from conceptual or calculation errors.

Choice B is incorrect and may result from conceptual or calculation errors.

Choice C is incorrect and may result from conceptual or calculation errors.

Choice B is correct. This probability is calculated by dividing the number of baritone singers by the total number of men registered for singing lessons. It’s given that a total of 25 men registered for singing lessons and that there are 10 baritones. Therefore, the probability of selecting a baritone from this group at random is , which is equivalent to 0.40.

Choice A is incorrect. This would be the probability of selecting a baritone at random if there were 100 total men who registered for singing lessons. Choice C is incorrect. This is the probability of selecting a singer at random who isn’t a baritone. Choice D is incorrect. This would be the probability of selecting a baritone at random if there were 15 total men registered for singing lessons.

Choice A is correct. It's given that an equation for the exponential model shown can be written as $y=a{\left(b\right)}^x$, where $a$ and $b$ are positive constants. For an exponential model written in this form, if the value of $b$ is greater than $0$ but less than $1$, the model is decreasing. If the value of $b$ is greater than $1$, the model is increasing. The exponential model shown is decreasing. Therefore, the value of $b$ is greater than $0$ but less than $1$. Of the given choices, only $0.83$ is a value greater than $0$ but less than $1$. Thus, $0.83$ is closest to the value of $b$.

Choice B is incorrect and may result from conceptual or calculation errors.

Choice C is incorrect and may result from conceptual or calculation errors.

Choice D is incorrect and may result from conceptual or calculation errors.

Choice B is correct. According to the table, the minimum wage in 1960 was $1.00 per hour, and in 1970 it was $1.60 per hour. The percentage change is therefore .

Choice A is incorrect and may result from averaging the two wages before calculating the percentage change. Choice C is incorrect. This is the 1960 wage expressed as a percentage of the 1970 wage, not the percentage change between the two. Choice D is incorrect and may result from a calculation error.

Choice A is correct. It's given that the bar graph shows the distribution of the number of students in each of four extracurricular activities at a high school. The bar representing drama has a height of $40$; therefore, there are $40$ students in drama. The bar representing chess has a height of $30$; therefore, there are $30$ students in chess. Thus, there are $40-30$, or $10$ more students in drama than in chess.

Choice B is incorrect. This is the number of students in chess.

Choice C is incorrect. This is the number of students in drama.

Choice D is incorrect. This is the sum of the number of students in drama and in chess.

Choice A is correct. It’s given that the ratio of black pens to red pens is $8$ to $1$. Therefore, there are $\frac{1}{8}$ as many red pens as black pens in the box. It’s also given that there are $40$ black pens in the box. Therefore, the number of red pens is $\frac{1}{8}$ of the $40$ black pens. Thus, the number of red pens is $40\left(\frac{1}{8}\right)$, or $5$.

Choice B is incorrect. This is the number of black pens in the box for every red pen.

Choice C is incorrect. This is the number of black pens in the box.

Choice D is incorrect. This is the number of red pens in the box if the ratio of black pens to red pens is $1$ to $8$.

Choice B is correct. The percent increase from 2012 to 2013 was , or 5%. Since the percent increase from 2012 to 2013 was estimated to be double the percent increase from 2013 to 2014, the percent increase from 2013 to 2014 was expected to be 2.5%.
Therefore, the number of subscriptions sold in 2014 is expected to be the number of subscriptions sold in 2013 multiplied by , or .

Choice A is incorrect and is the result of adding half of the value of the increase from 2012 to 2013 to the 2013 result. Choice C is incorrect and is the result adding twice the value of the increase from 2012 to 2013 to the 2013 result. Choice D is incorrect and is the result of interpreting the percent increase from 2013 to 2014 as double the percent increase from 2012 to 2013.

Choice C is correct. Because a random sample of the city council was polled, the proportion of the sample who supported the bill is expected to be approximately equal to the proportion of the total city council who supports the bill. Since 6 of the 20 polled, or 30%, supported the bill, it can be estimated that , or 15, city council members support the bill.

Choice A is incorrect. This is the number of city council members in the sample who supported the bill. Choice B is incorrect and may result from a computational error. Choice D is incorrect. This is the number of city council members in the sample of city council members who were not polled.

The correct answer is 4.44. The selling price, in dollars per cubic inch, is found by multiplying the density, in ounces per cubic inch, by the unit price, in dollars per ounce: yields . Thus, the selling price, in dollars per cubic inch, is 4.44.

Choice C is correct. It’s given that the population density of Worthington is $290$ people per square mile and Worthington has a population of $\mathrm{92,800}$ people. Therefore, the area of Worthington is $\mathrm{92,800} \text{people}\left(\frac{1 \text{square} \text{mile}}{290 \text{people}}\right)$, which is equivalent to $\frac{\mathrm{92,800} \text{square} \text{miles}}{290}$, or $320$ square miles.

Choice A is incorrect and may result from conceptual or calculation errors.

Choice B is incorrect and may result from conceptual or calculation errors.

Choice D is incorrect and may result from conceptual or calculation errors.

Choice C is correct. The mean of a data set is found by dividing the sum of the values in the data set by the number of values in the data set. Therefore, the mean of data set A is , which simplifies to 74. The mean of data set B is represented by the equation , or . It’s given that the mean of data set A is equal to the mean of data set B. Therefore, the equation can be used to solve for x. Multiplying both sides of this equation by 5 yields . Subtracting 284 from both sides of this equation yields .

Choices A, B, and D are incorrect and may result from calculation errors.

The correct answer is 75. The percent decrease between two values is found by dividing the difference between the two values by the original value and multiplying by 100. The line graph shows that the annual snowfall in 2003 was 40 inches, and the annual snowfall in 2007 was 10 inches. Therefore, the percent decrease in the annual snowfall from 2003 to 2007 is , or 75. It’s given that this is equivalent to , so the value of p is 75.

Choice A is correct. Dividing the number of revolutions by the number of minutes gives the number of revolutions the turbine completes per minute. It’s given that the wind turbine completes $900$ revolutions in $50$ minutes. Therefore, at this rate, this turbine completes $\frac{900}{50}$, or $18$, revolutions per minute.

Choice B is incorrect and may result from conceptual or calculation errors.

Choice C is incorrect and may result from conceptual or calculation errors.

Choice D is incorrect and may result from conceptual or calculation errors.

Choice C is correct. For the given line graph, the percent of cars for sale at a used car lot on a given day is represented on the vertical axis. The percent of cars for sale is the smallest when the height of the line graph is the lowest. The lowest height of the line graph occurs for cars with a model year of $2014$.

Choice A is incorrect and may result from conceptual errors.

Choice B is incorrect and may result from conceptual errors.

Choice D is incorrect and may result from conceptual errors.

Choice A is correct. It’s given that there are $\mathrm{2,358}$ raccoons in a $131$-square-mile area. The estimated population density, in raccoons per square mile, is the estimated number of raccoons divided by the number of square miles. Therefore, the estimated population density of this area is $\frac{\mathrm{2,358}\text{ raccoons}}{131 \text{square miles}}$, or $18$ raccoons per square mile.

Choice B is incorrect. This is the number of square miles in the area, not the estimated number of raccoons per square mile in this area.

Choice C is incorrect and may result from conceptual or calculation errors.

Choice D is incorrect and may result from conceptual or calculation errors.

Choice B is correct. The table summarizes all 1,000 responses from the students surveyed. If 312 are males who play a sport, 220 are females who play a sport, and 216 are females who do not play a sport, then 1,000 – 312 – 220 – 216 = 252 males who do not play a sport.

Choices A, C, and D are incorrect. If 109 males who do not play a sport responded, then the table summary would be 109 + 312 + 220 + 216 = 857 total student responses rather than 1,000. If 468 males who do not play a sport responded, then the table summary would be 468 + 312 + 220 + 216 = 1,216 total student responses rather than 1,000. If 688 males who do not play a sport responded, then the table summary would be 688 + 312 + 220 + 216 = 1,436 total student responses rather than 1,000.
 

Choice D is correct. The sample of 150 largemouth bass was selected at random from all the largemouth bass in the pond, and since 30% of the fish in the sample weighed more than 2 pounds, it can be concluded that approximately 30% of all largemouth bass in the pond weigh more than 2 pounds.

Choices A, B, and C are incorrect. Since the sample contained 150 largemouth bass, of which 30% weighed more than 2 pounds, this result can be generalized only to largemouth bass in the pond, not to all fish in the pond.

Choice D is correct. An appropriate model should follow the trend of the data points and should have data points both above and below the model. The scatterplot shows that the data points have an increasing trend that is curved. Therefore, an appropriate model should be an increasing curve with data points both above and below the model. Of the given choices, only the model in choice D is an increasing curve with data points both above and below the model.

Choice A is incorrect. Since the trend of the data points isn't linear, a line isn't the most appropriate model for the data.

Choice B is incorrect. Since the trend of the data points is increasing and isn't linear, a decreasing line isn't the most appropriate model for the data.

Choice C is incorrect. All the data points are below the model shown in this graph.

The correct answer is 8. In this group, of the people who are rhesus negative have blood type B. The total number of people who are rhesus negative in the group is , and there are 2 people who are rhesus negative with blood type B. Therefore, . Combining like terms on the left-hand side of the equation yields . Multiplying both sides of this equation by 9 yields , and multiplying both sides of this equation by yields . Subtracting 10 from both sides of this equation yields .

Choice A is correct. The total number of state parks in the state is . According to the table, 5 of these have camping facilities but not bicycle paths. Therefore, if a state park is selected at random, the probability that it has camping facilities but not bicycle paths is .

Choice B is incorrect. This is the probability that a state park selected at random from the state parks with camping facilities does not have bicycle paths. Choice C is incorrect. This is the probability that a state park selected at random from the state parks with bicycle paths does not have camping facilities. Choice D is incorrect. This is the probability that a state park selected at random from the state parks without bicycle paths does have camping facilities.

Choice D is correct. It’s given that there are $7$ red, $4$ white, $33$ blue, and $33$ yellow cubes in the bag. Therefore, there are a total of $7+4+33+33$, or $77$, cubes in the bag. If the cube is neither blue nor yellow, then it must be either red or white. Therefore, the probability of selecting a cube that is neither blue nor yellow is equivalent to the probability of selecting a cube that is either red or white. If one of these cubes is selected at random, the probability of selecting a cube that is either red or white is equal to the sum of the number of red cubes and white cubes divided by the total number of cubes in the bag. There are $7$ red cubes, $4$ white cubes, and $77$ total cubes in the bag. Therefore, the probability of selecting a red or white cube is $\frac{7+4}{77}$, which is equivalent to $\frac{11}{77}$, or $\frac{1}{7}$. Thus, if one cube is selected at random, the probability of selecting a cube that is neither blue nor yellow is $\frac{1}{7}$.

Choice A is incorrect. This is the probability of selecting a cube that is either blue or yellow, rather than the probability of selecting a cube that is neither blue nor yellow.

Choice B is incorrect and may result from conceptual or calculation errors.

Choice C is incorrect and may result from conceptual or calculation errors.

Choice A is correct. The average can be found by dividing the total number of boxes sold by the number of students, which is .

Choices B, C, and D are incorrect. Each results from choosing measures that require the results of individual students, which are not given.

The correct answer is $\frac{1}{2}$. For the graph shown, $x$ represents time, in seconds, and $y$ represents momentum, in newton-seconds. Therefore, the average rate of change, in newton-seconds per second, in the momentum of the object between two x-values is the difference in the corresponding y-values divided by the difference in the x-values. The graph shows that at $x=2$, the corresponding y-value is $6$. The graph also shows that at $x=6$, the corresponding y-value is $8$. It follows that the average rate of change, in newton-seconds per second, from $x=2$ to $x=6$ is $\frac{8-6}{6-2}$, which is equivalent to $\frac{2}{4}$, or $\frac{1}{2}$. Note that 1/2 and .5 are examples of ways to enter a correct answer.

Choice C is correct. Standard deviation is a measure of spread, so data sets with larger standard deviations tend to have larger spread. The standard deviation of the heights of the men in Group B is larger than the standard deviation of the heights of the men in Group A. Therefore, the heights of the men in Group B had a larger spread than the heights of the men in Group A.

Choice A is incorrect. If two data sets are identical, they will have equivalent means and equivalent standard deviations. Since the two data sets have different standard deviations, they cannot be identical. Choice B is incorrect. Without knowing the maximum value for each data set, it’s impossible to know which group contained the tallest participant. Choice D is incorrect. Since the means of the two groups are equivalent, the medians could also be the same or could be different, but it's impossible to tell from the given information.

Choice D is correct. It's given that $a$ is $230\%$ of $b$. It follows that $a=\frac{230}{100}b$. It's also given that $a$ is $60\%$ of $c$. It follows that $a=\frac{60}{100}c$. Since $a=\frac{230}{100}b$ and $a=\frac{60}{100}c$, it follows that $\frac{230}{100}b=\frac{60}{100}c$. Multiplying each side of this equation by $\frac{100}{60}$ yields $\frac{23}{6}b=c$. If $c$ is $p\%$ of $b$, then $c=\frac{p}{100}b$. It follows that $\frac{23}{6}=\frac{p}{100}$. Multiplying each side of this equation by $100$ yields $\frac{\mathrm{2,300}}{6}=p$. It follows that the value of $p$ is approximately $383.33$. Therefore, of the given choices, $383$ is closest to the value of $p$.

Choice A is incorrect. This is closest to the value of $p$ if $b$ is $230\%$ of $a$, rather than if $a$ is $230\%$ of $b$, and if $b$ is $p\%$ of $c$, rather than if $c$ is $p\%$ of $b$.

Choice B is incorrect. This is closest to the value of $p$ if $a$ is $230\%$ greater than $b$, rather than $230\%$ of $b$.

Choice C is incorrect. This is closest to the value of $p$ if $c$ is $p\%$ greater than $b$, rather than $p\%$ of $b$.

Choice B is correct. The height of each bar in the graph shown represents the number of containers that contain the number of walnuts specified at the bottom of the bar. The bar for $78$ walnuts has a height of $7$. Therefore, $7$ of these containers of walnuts contain exactly $78$ walnuts.

Choice A is incorrect. This is the number of containers that contain exactly $75$ walnuts, not $78$ walnuts.

Choice C is incorrect. This is the total number of containers of walnuts represented in the bar graph, not the number that contain exactly $78$ walnuts.

Choice D is incorrect. This is the number of walnuts in a container that contains exactly $78$ walnuts, not the number of containers that contain exactly $78$ walnuts.

Choice A is correct. If x is 6, then the five numbers in the given list are 15, 14, 18, 17, 6. The mean of these five numbers is the sum of all the values divided by the number of values, or . The median of these five numbers can be found by ordering the numbers from least to greatest and determining the middle value. When ordered from least to greatest, the numbers in the given list are 6, 14, 15, 17, 18, and the middle value is 15. Since the mean is 14 and the median is 15, the mean and median aren’t equal when x is 6.

Choices B, C, and D are incorrect. If any of these values is substituted for x, the mean and median of the data set would be equal.

Choice B is correct. It's given that a data set consists of $173$ colors and the bar graph shows the number of times each color appears in the data set. Therefore, for each color specified at the bottom of the bar, the frequency corresponds to the number of times that color appears in the data set. The color that appears $70$ times in the data set has a frequency of $70$ on the bar graph. Since the bar with a frequency of $70$ corresponds to green, green is the color that appears $70$ times.

Choice A is incorrect. The color blue appears about $27$ times, not $70$ times.

Choice C is incorrect. The color red appears about $33$ times, not $70$ times.

Choice D is incorrect. The color yellow appears about $43$ times, not $70$ times.

Choice A is correct. Let $x$ represent the percentage of $300$ that is $75$. This can be written as $\frac{x}{100}\left(300\right)=75$, or $3x=75$. Dividing both sides of this equation by $3$ yields $x=25$. Therefore, $25\%$ of $300$ is $75$.

Choice B is incorrect. $50\%$ of $300$ is $150$, not $75$.

Choice C is incorrect. $75\%$ of $300$ is $225$, not $75$.

Choice D is incorrect. $225\%$ of $300$ is $675$, not $75$.

The correct answer is $52$. The mean of a data set is calculated by dividing the sum of the data values by the number of values. It’s given that data set A consists of $10$ values, $9$ of which are shown. Let $x$ represent the $10\text{th}$ data value in data set A, which isn’t shown. The mean of data set A can be found using the expression $\frac{43+45+44+43+38+39+40+46+40+x}{10}$, or $\frac{378+x}{10}$. It’s given that the mean of the $9$ values shown is $42$ and that the mean of all $10$ numbers is greater than $42$. Consequently, the $10\text{th}$ data value, $x$, is larger than $42$. It’s also given that the data values in data set A are positive integers less than $60$. Thus, $42<x<60$. Finally, it’s given that the mean of data set A is an integer. This means that the sum of the $10$ data values, $378+x$, is divisible by $10$. Thus, $378+x$ must have a ones digit of $0$. It follows that $x$ must have a ones digit of $2$. Since $42<x<60$ and $x$ has a ones digit of $2$, the only possible value of $x$ is $52$. Since $52$ is larger than any of the integers shown, the largest integer from data set A is $52$.

Choice D is correct. Since the ratio of y to x is constant for each ordered pair in the table, the first row can be used to determine that the ratio of y to x is 4 to 1. The proportion can be used to solve for k. Multiplying each side of the equation by 40 yields .

Choice A is incorrect. This is the value of y when the value of x is 7, not 40. Choice B is incorrect and may result from subtracting 4 from 40 instead of multiplying 40 by 4. Choice C is incorrect and may result from incorrectly setting up the proportion.

Choice C is correct. The total cost to purchase the shirt is the $20 cost of the shirt plus the 5% sales tax. The value of the 5% sales tax on the $20 shirt is equivalent to, or $1. Therefore, the total cost to purchase the shirt is , or $21.

Choice A is incorrect and may result from neglecting to multiply by $20 when finding the value of the sales tax. Choice B is incorrect and may result from dividing by 10, instead of by 100, and then neglecting to multiply by $20 when finding the sales tax. Choice D is incorrect and may result from interpreting the sales tax of 5% as $5.